TEST 6 Review
1)
25+3x = 1/16
25+3x = 0.0625
log2[25+3x] = log2[0.0625]
Take base-2 log on both sides.
[5+3x]log2(2) = log10(0.0625)/log10(2)
Use Power Rule and change-of-base formula
5+3x = log10(0.0625)/log10(2)
Note: log2(2) = 1
5+3x = -4
5+3x - 5 = -4 – 5
3x = -9
x = -3
2) y = 4(4x-2)
3) y = 2x
4) (4/5)x = 16/25
0.8x = 0.64
log0.8[0.8x] = log0.8[0.64]
xlog0.8(0.8) = log10(0.64)/log10(0.8)
Take base-0.8 log of both sides.
Use Power Rule and change-of-base formula
x = log10(0.64)/log10(0.8)
Note: log0.8(0.8) = 1
x=2
5) 41+2x = 64
log4[41+2x] = log4[64]
[1+2x]log4(4) = log10(64)/log10(4)
Take base-4 log of both sides.
Use Power Rule and change-of-base formula
1+2x = log10(64)/log10(4)
Note: log4(4) = 1
1+2x = 3
1+2x - 1= 3 – 1
2x = 2
X=1
6) f(x) = ax
If a is allowed to be negative, the function could not be defined for
all values of x, such as negative values.)
7) y = (1/2)x
8) 4x = 1/16
4x = 0.015625
log4[4x] = log4[0.015625]
xlog4(4) = log10(0.015625)/log10(4)
Take base-4 log of both sides.
Use Power Rule and change-of-base formula
x = log10(0.015625)/log10(4)
x = -3
9) Write 3 27  3 in log form
Note:
3
3
27  3 271  271/ 3
27  3
271/ 3  3
log 27 3  1/ 3
10) Log4(7)
Note: log4(4) = 1
When t = 0, f(t) = 8
12) log5(1/25) = -2
5-2 = 1/25
13) 62 = 36
log6(36) = 2
14) Evaluate log8(1/512)
= log10(1/512) / log10(8)
= -3
 1 
log10  
 1 
 25   2
15) Evaluate log 5 


25
log10  5 
 
16) Write 43 = 64 in log form
log4(64) = 3
17) Write logt(t) - logt(s) + 3 logt(u)
logt(t) - logt(s) + logt(u)3
log t (t) + log t (u)3 - log t (s)
log t (t  u 3 ) - log t (s)
 t  u3 
log t 

 s 
as a single logarithm.
8
18) Write log 5
x9 y
as sum and/or difference of logarithms.
z2
x1/8  y1/ 9
log 5
z2
log 5 x1/8  log 5 y1/ 9  log 5 z 2
1
1
log 5 x  log 5 y  2log 5 z
8
9
19) Evaluate 2
log 2  49 
log property: log b  b x   x
log property: b
Thus, 2
log 2  49 
log b  x 
 49
x
20) Write log n 6
9 x2
as sum and/or difference of logarithms.
3
z
1
 9 x2 
log n 6  3 
 z 
1/ 6
 9 x2 
log n  3 
 z 
 9 x2 
1
log n  3 
6
 z 
1
log n 9  log n x 2 - log n z 3 
6
1
log n 9  2log n x - 3log n z 
6
1
2
3
log n 9  log n x   log n z 
6
6
6
1
1
1
log n 9  log n x   log n z 
6
3
2
21)Write log m m  log m n as a single logarithm.
log m  m  n 
22)Write log12
17 x
as sum and/or difference of logarithms.
y
17 x1/ 2
log12
y
log12 17  log12 x1/ 2  log12 y
log12 17  12 log12 x  log12 y
3
1
23) Write log a x  log a y  log a w  6log a z as a single logarithm.
5
2
log a x  log a y 3/ 5  log a w1/ 2  log a z 6
log a x  log a w1/ 2  log a y 3/ 5  log a z 6
 log x  log w    log y +log z 
 log x  w    log y  z 
1/ 2
a
3/ 5
a
a
1/ 2
a
 x  w1/ 2 
log a  3/ 5 6 
 y z 
3/ 5
a
6
a
6
x2 y5
24)Write log 4
as sum and/or difference of logarithms.
7
log 4 x 2  log 4 y 5  log 4 7
2log 4 x  5log 4 y  log 4 7
25) ln 0.986 = -0.0140989243795
26) log(0.00314)
log10 (0.00314)  2.503070351926
27) Solve log 6 ( x  1)  log 6 ( x  4)  2
Note: 2 written in terms of log base-6 is
2 = log 6 (62 )  log 6 (36)
log 6 ( x  1)  log 6 ( x  4)  2
log 6 ( x  1)  log 6 ( x  4)  log 6 (36)
log 6  ( x  1)  ( x  4)   log 6 (36)
( x  1)  ( x  4)  36
x 2  4 x  x  4  36
x 2  3 x  4  36
x 2  3 x  4  36  36  36
x 2  3 x  40  0
x 2  3 x  40  0
( x  8)( x  5)  0
set x + 8 = 0
x = -8
set x - 5 = 0
x=5
Check answers:
For x = -8
log 6 ( x  1)  log 6 ( x  4)  2
log 6 ( 8  1)  log 6 ( 8  4)  2
log 6 ( 9)  log 6 ( 4)  2
Since log of a negative number is undefined,
-8 is an extraneous solution.
For x = 5
log 6 ( x  1)  log 6 ( x  4)  2
log 6 (5  1)  log 6 (5  4)  2
log 6 (4)  log 6 (9)  2
log 6 (4  9)  2
log 6 (36)  2
22
Solution set is {5}
28) Solve log 5 ( x  1)  3  log 5 (6 x  4)
Note: 3 written in terms of log base-5 is
3 = log 5 (53 )  log 5 (125)
log 5 ( x  1)  log 5 (125)  log 5 (6 x  4)
log 5 ( x  1)  log 5 (125)  log 5 (6 x  4)
 x  1
log 5 
 log 5 (6 x  4)

125


x 1
 6x  4
125
x 1
(125)
  6 x  4  125
125
x  1  750 x  500
x  1   x  1  750 x  500   x  1
0  750 x  500  x  1
0  749 x  501
0  501  749 x  501  501
501  749 x
501 749 x

749
749
501
x
749
Check Answer:
501
x
749
log 5 ( x  1)  3  log 5 (6 x  4)
  501  
  501 

log 5   

1

3

log
6


4
5 
 


  749  
  749 

  501 

log 5  negative number   3  log 5  6  

4


  749 

501
x
is an extraneous solution.
749
29) ln(4.37 102 )  ln(0.0437)  3.130407176
30) log 2.18  log10 2.18  0.3384564936
31)
193-x = 26
log19[193-x] = log19[26]
[3-x]log19(19) = log10(26)/log10(19)
3-x = log10(26)/log10(19)
Take base-19 log of both sides.
Use Power Rule and change-of-base formula
Note: log19(19) = 1
3-x = 1.10652540639298
3-x-3= 1.10652540639298 -3
-x = -1.89347459360702
X = 1.89347459360702
32)
6x+1 = 29
log6[6x+1] = log6[29]
[x+1]log6(6) = log10(29)/log10(6)
x+1 = log10(29)/log10(6)
Take base-6 log of both sides.
Use Power Rule and change-of-base formula
Note: log6(6) = 1
x+1 = 1.87932358545715
x+1-1 = 1.87932358545715 -1
x = 0.87932358545715;
33)
20x = 54
log20[20x] = log20[54]
xlog20(20) = log10(54)/log10(20)
x = log10(54)/log10(20)
x = 1.33155558718601
Take base-20 log of both sides.
Use Power Rule and change-of-base formula
Note: log20(20) = 1
34)
log₄(x²) = log₄(5x + 14)
(x²) = (5x+14)
x² = 5x + 14
x² - (5x + 14) = 5x + 14 - (5x + 14)
x²-5x - 14 = 0
(x-7)(x+2) = 0
Set x-7 = 0
set x+2 = 0
X=7
x = -2
Check answers:
log₄(x²) = log₄(5x + 14)
log₄((7)²) = log₄(5(7) + 14)
log₄(49) = log₄(49)
log₄((-2)²) = log₄(5(-2) + 14)
log₄(4) = log₄(4)
Solution set is {7, -2}
35)
log3(x) = 5
Note: 5 can be written as 5 = log3(35) = log3(243)
Rewrite log3(x) = 5 as follows:
log3(x) = log3(243)
x = 243
Solution set is {243}
Check answer:
log3(x) = 5
log3(243) = 5
5=5